The path difference between two waves $y_{1} = a_{1}sin\left( \omega t - \frac{2\pi x}{\lambda} \right)$ and $y_{2} = a_{2}\cos\left( \omega t - \frac{2\pi x}{\lambda} + \phi \right)$is

Question

Dr. Abubacker · Accepted Answer

(b) $y_{2} = a_{2}\cos\left( \omega t - \frac{2\pi x}{\lambda} + \phi ight)$
$= a_{2}\sin\left\lbrack \frac{\pi}{2} + \left( \omega t - \frac{2\pi x}{\lambda} + \phi ight) ightbrack$
Compare it with $y_{1} = a_{1}sin\left( \omega t - \frac{2\pi x}{\lambda} ight)$
Phase difference$= \left( \frac{\pi}{2} + \phi ight)$
$	herefore\ $Path difference $= \frac{\lambda}{2\pi}\left( \frac{\pi}{2} + \phi ight)$

Q. The path difference between two waves $y_{1} = a_{1}sin\left( \omega t - \frac{2\pi x}{\lambda} \right)$ and $y_{2} = a_{2}\cos\left( \omega t - \frac{2\pi x}{\lambda} + \phi \right)$is

Explanation

Reviewed by Dr. Abubacker

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